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\topic{Lecture 8 \\Multiple Integral\\ \scriptsize Triple Integration (30 Sep 2009)}
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\section{Triple Integral} 
Consider the function $f(x,y,z)$, which is given for all $x, y, z$ in a region $V$ of the three dimensional space. We devide the region $V$ by drawing the parallel lines to the $x, y$ and $z$ axes into $n$ elementary volumes. 

Let $(x_k,y_k,z_k)$ be any point inside the $k$th volume element, and $\Delta V_k$ is the volume of the $k$th element, then we form the sum 
\begin{equation}
I_k = \sum_{k=1}^{n}f(x_k,y_k,z_k).\Delta V_k
\end{equation}
Now as we increase the number of divisions taking smaller and smaller elementary volumes. In this fashion , as $n$ tends to infinity and the dimensions of each subdivision tends to zero, in such case we obtain a sequence of real numbers $I_{n_1}, I_{n_2},...$. Then The limit of this sequence is called the \textbf{triple integral} of $f(x,y,z)$ over the region $V$, and is denoted by the symbol
\[
\int\int\int_Vf(x,y,z)dxdydz
\]
Thus,
\begin{equation}
\label{eq:DIDef}
I=\int\int\int_Vf(x,y,z)dxdydz = \lim_{n\longrightarrow \infty} \sum_{k=1}^{n}f(x_k,y_k,z_k).\Delta V_k
\end{equation}

This definition is similar to that of a double integral. For the purpose of evaluation the triple integral also can be expressed as a repeated  integral.
\begin{example}
Evaluate the triple integral of the function $x^2$ over the region $V$ enclosed by the planes $x=0$, $y=0$, $z=0$, and $x+y+z=a$
changing the variables to polar coordinates.
\end{example}

Here first we take a vertical column with rectangular base on $xy$-plane, and other end on plane $x+y+z=a$. i.e. vertical column is bounded by the planes $z=0$ and $z=a-x-y$.Thus
\[ 0 \leq z \leq a-x-y\]
Now, look at $xy$-plane, the variation of base rectangle produces the limits
\[0 \leq y \leq a-x,~~~~0 \leq x \leq a\]
Hence the problem is now, the triple integral
\[
\begin{array}[2]{ll}
=  & \int_0^a\int_0^{a-x}\int_0^{a-x-y}x^2dxdydz \\
=  & \int_0^a\int_0^{a-x}x^2 [z]_0^{a-x-y} dxdy = \int_0^a\int_0^{a-x}x^2 [a-x-y] dxdy\\
=  & \int_0^a [ax^2y-x^3y-\frac{1}{2} x^2y^2]_0^{a-x} dxdy \\
= & \int_0^a [ax^2(a-x)-x^3(a-x)- \frac{1}{2} x^2(a-x)^2] dxdy =  \frac{1}{2} \int_0^a  x^2(a-x)^2] dxdy \\
= &  \frac{1}{2}\left[  \frac{1}{3}a^2x^3 -  \frac{2}{4}ax^4 +  \frac{1}{5}x^5 \right]_0^a =  \frac{1}{60}a^5
\end{array}
\]
\section*{Problems}
\begin{enumerate}
\item 	Evaluate $\int \int \int _{R} (x+y+z)  $ dx dy dz, where R:0 $\le x\le 1,1\le y\le 2,2\le z\le 3$.
[Ans -$\frac{9}{2}$]
\item Evaluate $\int \int \int _{R} (x-2y+z)  $dx dy dz, where R: 0$\le x\le 1,0\le y\le x^{2} ,0\le z\le x+y$.
[Ans - $\frac{8}{35}$]
\item Evaluate the integral: $\int _{0}^{\log 2}\int _{0}^{x}\int _{0}^{x+\log y}e^{x+y+z}dz dy dx$.
[Ans - $\frac{8}{3} \log 2-\frac{19}{9} $]
\item Evaluate $\int \int \int _{R} (x^{2} +y^{2} +z^{2} ) dx dy dz$, where $R$ denotes the region bounded by $x = 0, y = 0, z = 0$ and $x +y +z = a$, $(a>0)$. [Ans- $\frac{a^{5} }{20} $]
\item Compute $\int \int \int \frac{dxdydz}{(x+y+z+l)^{3} }$ if the region of integration is bounded by the co-ordinate planes and the plane $x + y + z =1$. [Ans - $\frac{1}{2} \log 2-\frac{5}{16} $]
\item Evaluate $\int \int \int _{S} xyz  $ dx dy dz where S =$\left[(x,y,z):x^{2} +y^{2} +z^{2} \le l,x\ge 0,y\ge 0,z\ge 0\right].$ 
[Ans - $\frac{1}{48}$]
\item Evaluate $\int \int \int _{S} \sqrt{x^{2} +y^{2} } dx dy dz $, where $S$ is the solid bounded by the surfaces $x^{2} +y^{2} =z^{2} ,z=0,z=1$.
[Ans - $\frac{\pi }{6} $].
\item Evaluate $\int _{0}^{2x} \int _{0}^{x/4} \int _{0}^{a} r^{2} $sin $\theta $ dr d$\theta $ d$\phi$.
[Ans - $\frac{2\pi a^{3} }{3\sqrt{2} } (\sqrt{2}-1)$]
\item Evaluate the integral $\int \int \int    $(x$^{2} +y^{2} +z^{2)} dx dy dz $ taken over the volume enclosed by the sphere x$^{2} +y^{2} +z^{2} =1.$
[Ans -$\frac{4\pi }{5} $]
\item Calculate the volume of the solid bounded by the surface $x = 0, y = 0 , x + y + z = 1$, and  $z = 0$. [Ans - $\frac{1}{6} $] 
\end{enumerate}
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